Maple Questions and Posts

These are Posts and Questions associated with the product, Maple

I have a differential equation involving several functions of the following form:

diff(h,z) = iAf + iBg,

where h, f and g are functions of the Cartesian coordinates x, y and R and the third coordinate corresponds to z = R for some fixed constant value R.  The derivative is then with respect to the coordinate z and A and B are constants, with i the usual imaginary unit.  Is there some way this equation could be solved explicitly with Maple?

Basically it spits out the subset of values for which a division by zero error will occur for the function you specify on  range you specify for each of it's arguments, but I get an ambigous error when ever exponentiation features in the function I specify, which of course dramatically reduces the application of the calculator. Division,addition,substraction and multiplication are currently the only available arithmetic operators availble for the function window that I know the error will not occur.

If some one can help it is much appriciated

 

DIVISION_BY_ZERO_CALCULATOR.mw

hello everyone,
   INGT.mw
 

L__d := 100:

L__b := 200:

L__c := L__d+(1/2)*L__b;

200

(1)

X := .3;

.3

(2)

Error, (in plot) procedure expected, as range contains no plotting variable

 

`ΔE__g` := 1.155*X+.37*X^2;

.3798

(3)

V__0 := .6*`ΔE__g`;

.22788

(4)

m__D := 0.67e-1*m[e];

0.67e-1*m[e]

(5)

m__B := (0.67e-1+0.83e-1*X)*m[e];

0.919e-1*m[e]

(6)

Sol := solve(2*cos(L__d*sqrt(E__x))+(m__D*sqrt((V__0-E__x)/E__x)/m__B-m__B*sqrt(E__x/(V__0-E__x))/m__D)*sin(L__d*sqrt(E__x))-(m__D*sqrt((V__0-E__x)/E__x)/m__B+m__B*sqrt(E__x/(V__0-E__x))/m__D)*sin(L__d*sqrt(E__x))*exp(-sqrt(V__0-E__x)*L__b) = 0, E__x);

0.1110897170e-2, 0.3531161505e-2, -.2585338615+0.9991335677e-27*I

(7)

 

0.1110897170e-2, 0.3531161505e-2, -.2585338615+0.9991335677e-27*I

(8)

E__1 := 0.1110897170e-2:

K__1 := sqrt(E__1);

0.3333012406e-1

(9)

K__2 := sqrt(V__0-E__1);

.4762027959

(10)

C := cosh((1/2)*K__2*L__b);

0.2399908351e21

(11)

beta := m__D*K__2/(m__B*K__1);

10.41631973

(12)

B := -beta*sinh((1/2)*K__2*L__b);

-0.2499821271e22

(13)

A := -B*sin(K__1*L__d)+C*cos(K__1*L__d);

-0.7112056933e21

(14)

h := proc (x) options operator, arrow; piecewise(x <= -L__c, A*exp(K__2*(x+L__c)), -L__c < x and x < -(1/2)*L__b, -B*sin(K__1*(x+(1/2)*L__b))+C*cos(K__1*(x+(1/2)*L__b)), abs(x) <= (1/2)*L__b, (1/2)*exp(K__2*x)+(1/2)*exp(-K__2*x), (1/2)*L__b < x and x < L__c, -B*sin(K__1*(x-(1/2)*L__b))+C*cos(K__1*(x-(1/2)*L__b)), L__c <= x, A*exp(K__2*(x-L__c))) end proc:

'h(x)' = h(x);

h(x) = piecewise(x <= -200, -7.112056933*10^20*exp(95.24055918+.4762027959*x), -200 < x and x < -100, 2.499821271*10^21*sin(3.333012406+0.3333012406e-1*x)+2.399908351*10^20*cos(3.333012406+0.3333012406e-1*x), abs(x) <= 100, (1/2)*exp(.4762027959*x)+(1/2)*exp(-.4762027959*x), 100 < x and x < 200, 2.499821271*10^21*sin(0.3333012406e-1*x-3.333012406)+2.399908351*10^20*cos(0.3333012406e-1*x-3.333012406), 200 <= x, -7.112056933*10^20*exp(-95.24055918+.4762027959*x))

(15)

L__y := 200:

L__z := 200:

P := proc (x, y, z) options operator, arrow; h(x)*cos(Pi*y/L__y)*cos(Pi*z/L__z) end proc:

'Psi(x, y, z)' = P(x, y, z);

Psi(x, y, z) = piecewise(x <= -200, -7.112056933*10^20*exp(95.24055918+.4762027959*x), -200 < x and x < -100, 2.499821271*10^21*sin(3.333012406+0.3333012406e-1*x)+2.399908351*10^20*cos(3.333012406+0.3333012406e-1*x), abs(x) <= 100, (1/2)*exp(.4762027959*x)+(1/2)*exp(-.4762027959*x), 100 < x and x < 200, 2.499821271*10^21*sin(0.3333012406e-1*x-3.333012406)+2.399908351*10^20*cos(0.3333012406e-1*x-3.333012406), 200 <= x, -7.112056933*10^20*exp(-95.24055918+.4762027959*x))*cos((1/200)*Pi*y)*cos((1/200)*Pi*z)

(16)

INGT := proc (x__i) `assuming`([evalf(int(int(int(P(x, y, z)^2*exp(-lambda*sqrt((x-x__i)^2+y^2+z^2)), x = -infinity .. infinity), y = -L__y .. L__y), z = -L__z .. L__z))], [0 < lambda]) end proc

evalf(INGT(2))

``

Warning,  computation interrupted

 

``


 

Download INGT.mw
 

L__d := 100:

L__b := 200:

L__c := L__d+(1/2)*L__b;

200

(1)

X := .3;

.3

(2)

Error, (in plot) procedure expected, as range contains no plotting variable

 

`&Delta;E__g` := 1.155*X+.37*X^2;

.3798

(3)

V__0 := .6*`&Delta;E__g`;

.22788

(4)

m__D := 0.67e-1*m[e];

0.67e-1*m[e]

(5)

m__B := (0.67e-1+0.83e-1*X)*m[e];

0.919e-1*m[e]

(6)

Sol := solve(2*cos(L__d*sqrt(E__x))+(m__D*sqrt((V__0-E__x)/E__x)/m__B-m__B*sqrt(E__x/(V__0-E__x))/m__D)*sin(L__d*sqrt(E__x))-(m__D*sqrt((V__0-E__x)/E__x)/m__B+m__B*sqrt(E__x/(V__0-E__x))/m__D)*sin(L__d*sqrt(E__x))*exp(-sqrt(V__0-E__x)*L__b) = 0, E__x);

0.1110897170e-2, 0.3531161505e-2, -.2585338615+0.9991335677e-27*I

(7)

 

0.1110897170e-2, 0.3531161505e-2, -.2585338615+0.9991335677e-27*I

(8)

E__1 := 0.1110897170e-2:

K__1 := sqrt(E__1);

0.3333012406e-1

(9)

K__2 := sqrt(V__0-E__1);

.4762027959

(10)

C := cosh((1/2)*K__2*L__b);

0.2399908351e21

(11)

beta := m__D*K__2/(m__B*K__1);

10.41631973

(12)

B := -beta*sinh((1/2)*K__2*L__b);

-0.2499821271e22

(13)

A := -B*sin(K__1*L__d)+C*cos(K__1*L__d);

-0.7112056933e21

(14)

h := proc (x) options operator, arrow; piecewise(x <= -L__c, A*exp(K__2*(x+L__c)), -L__c < x and x < -(1/2)*L__b, -B*sin(K__1*(x+(1/2)*L__b))+C*cos(K__1*(x+(1/2)*L__b)), abs(x) <= (1/2)*L__b, (1/2)*exp(K__2*x)+(1/2)*exp(-K__2*x), (1/2)*L__b < x and x < L__c, -B*sin(K__1*(x-(1/2)*L__b))+C*cos(K__1*(x-(1/2)*L__b)), L__c <= x, A*exp(K__2*(x-L__c))) end proc:

'h(x)' = h(x);

h(x) = piecewise(x <= -200, -7.112056933*10^20*exp(95.24055918+.4762027959*x), -200 < x and x < -100, 2.499821271*10^21*sin(3.333012406+0.3333012406e-1*x)+2.399908351*10^20*cos(3.333012406+0.3333012406e-1*x), abs(x) <= 100, (1/2)*exp(.4762027959*x)+(1/2)*exp(-.4762027959*x), 100 < x and x < 200, 2.499821271*10^21*sin(0.3333012406e-1*x-3.333012406)+2.399908351*10^20*cos(0.3333012406e-1*x-3.333012406), 200 <= x, -7.112056933*10^20*exp(-95.24055918+.4762027959*x))

(15)

L__y := 200:

L__z := 200:

P := proc (x, y, z) options operator, arrow; h(x)*cos(Pi*y/L__y)*cos(Pi*z/L__z) end proc:

'Psi(x, y, z)' = P(x, y, z);

Psi(x, y, z) = piecewise(x <= -200, -7.112056933*10^20*exp(95.24055918+.4762027959*x), -200 < x and x < -100, 2.499821271*10^21*sin(3.333012406+0.3333012406e-1*x)+2.399908351*10^20*cos(3.333012406+0.3333012406e-1*x), abs(x) <= 100, (1/2)*exp(.4762027959*x)+(1/2)*exp(-.4762027959*x), 100 < x and x < 200, 2.499821271*10^21*sin(0.3333012406e-1*x-3.333012406)+2.399908351*10^20*cos(0.3333012406e-1*x-3.333012406), 200 <= x, -7.112056933*10^20*exp(-95.24055918+.4762027959*x))*cos((1/200)*Pi*y)*cos((1/200)*Pi*z)

(16)

INGT := proc (x__i) `assuming`([evalf(int(int(int(P(x, y, z)^2*exp(-lambda*sqrt((x-x__i)^2+y^2+z^2)), x = -infinity .. infinity), y = -L__y .. L__y), z = -L__z .. L__z))], [0 < lambda]) end proc

evalf(INGT(2))

``

Warning,  computation interrupted

 

``


 

Download INGT.mw

 

I'm trying to calculate a triple integral complicated by a procedure that changes each time a variable xi, while the program takes a lot of time and it gives me the message "Warning, computation interrupted". If anyone can help me I will be very happy

Is there something I should be doing whenever I use simplify to avoid things like this, or should I stop using the "is" function all together?

 

interface(showassumed = 0):

 

sum(binomial(k+j, k), j = 0 .. n-k) = binomial(n+1, k+1)

(n-k+1)*binomial(n+1, k)/(k+1) = binomial(n+1, k+1)

(1)

#And we have:
is(sum(binomial(k+j, k), j = 0 .. n-k) = binomial(n+1, k+1))

FAIL

(2)

#And since:
is(simplify(convert(sum(binomial(k+j, k), j = 0 .. n-k) = binomial(n+1, k+1), 'factorial')))

true

(3)

is(sum(binomial(k+j, k), j = 0 .. n-k) = binomial(n+1, k+1)) = is(simplify(convert(sum(binomial(k+j, k), j = 0 .. n-k) = binomial(n+1, k+1), 'factorial')))


 

Download main.mw

Hello people :) 

As the captian says, im trying to remove an old task i've made.
But i get this:

Error in Get, invalid object [_XML_reply_data_get("reference" =
"_Maplets_reference_12","parameter" =
"value",_XML_content("Task,UserTasks,Nyops",&Entity "#xc3",&Enity
"#xa6","tning"))]

And i have no idea what it is, but it won't erase my task :'D

Thanks a bunch in advance! 

Have a great weekend you all
Best regards Lucas :)

A few seconds after calling up Help starts zucking araound and the whole computer then freezes. Ctrl-Alt-Delete doesn't work, hard reset required. Very funny. Am I alone?

Hi good afternoon , im looking for maple code of HPM to solve time dependent diffusion reaction , can anybody help me . I didnt get how to solve it by maple . 

Hello

I know that anames( 'user' ) recovers the contents of the workspace but I wonder whether it would be possible to recover only the contents of each m-file read within a worksheet.  

Many thanks

Ed

 

how to convert a function to transition matrix?

how to convert a differential system to transition matrix?

 

 

 

In addition to defining a function in a standard way, a fast and convenient way to turn a formula (an expression) into a function is to use unapply command (check Maple help center for this command). The following items are about defining functions. 

 

 

  1. Define a function midpoint, which returns the average of two arguments given on input. For example, midpoint(2,3) returns 2.5, midpoint(a,b) returns 𝑎+𝑏2. 
  2. b) For two positive natural numbers 𝑎 and 𝑏 we can define the least common multiple lcm(𝑎,𝑏) via their greatest common divisor gcd (𝑎,𝑏), i.e.: lcm(𝑎,𝑏) = 𝑎𝑏gcd (𝑎,𝑏). When 𝑎 and/or 𝑏 are zero, lcm(𝑎; 𝑏)=0. Use the arrow operator and piecewise to define the function my_lcm which returns the least common multiple of two natural numbers. 
  3. c) The quadratic formula allows to write the roots of 𝑝 = 𝑎𝑥2+𝑏𝑥+𝑐 explicitly in terms of 𝑎,𝑏, and 𝑐. 

 

 Define a piecewise function of 𝑥 so that the value of the function is 0, if 𝑥<−(𝑟1+ 𝑟2+𝑟3) or 𝑥>𝑟1+ 𝑟2+𝑟3, and it is equal to the area of confined region if −(𝑟1+ 𝑟2+𝑟3)<𝑥<𝑟1+ 𝑟2+𝑟3. 

 

 

a) Given real values 𝑟1,𝑟2,𝑟3>0, consider the curve of the positive part of the circle with radius 𝑟1+ 𝑟2+𝑟3 and center at the origin and three semi-circles below with radii 𝑟1,𝑟2,𝑟3 so that they are arranged from left to right like the following graph. Define proper functions and plot this graph in Maple for 𝑟1=10,𝑟2=2,𝑟3=8. 

Hey everyone,

Got another Symbolic question here that I have no idea how to begin. Please help again.

Question: 

Note that if i is an integer then convert(i,base, 2) will convert i to a binary list (the binary representation of i in reverse order). Then you may use list_to_set to convert  this list to a set. Note that if i runs from 0 to 2^n-1 then list_to_set(convert(i,base,2) )  will run through all subsets of {1, 2, . . ., n}. 

(a) Use this idea to make your own procedure PowerSet which will given n,  produce the powerset of {1, 2, . . . , n}. Show by several examples that your procedure works. For your examples you should get nops(PowerSet(n)) = 2^(n). Check that this is the case.
 
(b) Use this idea to make a procedure RandSet which given n produces a random subset of {1, 2, . . . , n}. [Use rand to produce a random integer in the range 0..2^n-1 and then convert it to a subset of {1, 2, . . . , n}.]  Do NOT use PowerSet or powerset. Show by examples that it works for small n such as 5, 10, and 20 as well as for large n such as 100. It will even work for n = 1000, but the output will be rather large. On average a random subset of {1,2,...,n} will contain n/2 elements.

Again, please help me with this and thank you in advance. Your assistance is appreciated.

Hi all,

First-time poster here. Got a question for Symbolic Computations and don't know how to do it. Please help me out.

Here is the question: 

There is a one-to-one correspondence between subsets of {1, 2, . . . , n} and binary lists of length n, that is, lists L = [x1, x2 , . . . , xn] where x1, x2, . . . , xn are elements of the set {0,1}.  The correspondence is given by associating to the set S the list L where xi = 1 if i is in S and 0 if not. For example, the set {1,3,5} corresponds to the list [1,0,1,0,1,0,0] if n = 7.

(a) Write a procedure list_to_set whose input is a binary list and whose output is the corresponding set. E. g., list_to_set([1,0,1,0,1]) will return the set {1,3,5}. Note that nops(L) is the length of a list.

(b) Write a procedure set_to_list whose input is a pair S,n where S is a subset of {1, 2, . . . , n} and n is a positive integer and whose output is the binary list of length n corresponding to the set S. E. g., if n = 5 then set_to_list({1,3,5},5) will return [1,0,1,0,1].

(c) Show by a few tests that each procedure works. Then apply set_to_list to each set in the powerset of {1, 2, 3, 4} to form all binary lists of length 4. Make a program to print out a table of the following form. (But the order need not be the same as that started below.)

   [0,0,0,0] <-->  {  }
   [1,0,0 0] <--> { 1 }
   [0,1,0,0] <--> { 2 } 
    ........
    etc

Some extra commas in the output is okay. You may obtain the power set of the set {1,2,...,n} by the command powerset(n); but you must first load the package combinat.

Please let me know if there are any questions. Thank you in advance.

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